My teacher found this cool shortcut for factoring. I would like to use, for it saves time, but I feel hesitant using it without knowing the mathematical proof. Can anyone watch the video and explain it to me, thanks. The link to the video is here, http://m.youtube.com/watch?v=r1JAJfmRG5w.
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Start with any quadratic with integer coefficients that have no common factor and A > 0: $$Ax^2+Bx+C$$
Now let's follow what your teacher did.
Step 1. Write $x^2+Bx+AC$.
Step 2. Factor $x^2+Bx+AC=(x-q)(x-p)$.
Step 3. Write $(x-\frac{q}{A})(x-\frac{p}{A})$ and reduce the fractions $\frac{q}{A}=\frac{r}{s}$ and $\frac{p}{A}=\frac{t}{k}$.
Step 4. The desired factorizations is $(sx-r)(kx-t)$.
To prove this works, note from step 2 that $B=-(q+p)$ and $AC = pq$. This is sufficient information to obtain the roots of $Ax^2+Bx+C$ by the quadratic formula: $$ x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}= \frac{q+p\pm \sqrt{(p+q)^2-4pq}}{2A}= \frac{q+p\pm \sqrt{(q-p)^2}}{2A}.$$ So the roots are $\frac{q}{A}$ and $\frac{p}{A}$. This proves that the roots of the given trinomial are the same as the factored form obtained but quadratics that agree on roots are only the same up to a constant factor. Actually proving that the method also agrees with this constant factor is actually the tricky part.
This method fails if gcf(A,B,C) is not equal 1 or if A is negative. If one is factoring 2x^2+4x+2 this method has you consider y^2+4y+4 with q=p=-2 then q/A=p/A=-1 giving r=t=-1 and s=k=1 (or vice versa) but unfortunately (x+1)(x+1)=(x+1)^2 is not equal to the original trinomial though 2(x+1)^2 is the correct factorization. -1x^2+1 also fails leading to (x-1)(x+1) instead of -(x-1)(x+1). Since students are commonly instructed to factor out the gcf (or -1 in the second case) the flaw is not noticed. However the proof in incomplete.
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